Essentials of Pharmaceutical Chemistry ( PDFDrive ).pdf

Essentials of Pharmaceutical Chemistry Essentials of Pharmaceutical Chemistry FOURTH EDITION Edited by Donald Cairns BSc, PhD, MRPharmS, CSci, CChem, FRSC Acting Head of School of Pharmacy and Life Sciences, The Robert Gordon University, Aberdeen, UK...

Essentials of Pharmaceutical Chemistry Essentials of Pharmaceutical Chemistry FOURTH EDITION Edited by Donald Cairns BSc, PhD, MRPharmS, CSci, CChem, FRSC Acting Head of School of Pharmacy and Life Sciences, The Robert Gordon University, Aberdeen, UK London Chicago Dedication For Elaine, Andrew and Mairi Published by Pharmaceutical Press 1 Lambeth High Street, London SE1 7JN, UK c Royal Pharmaceutical Society of Great Britain 2012 is a trade mark of Pharmaceutical Press Pharmaceutical Press is the publishing division of the Royal Pharmaceutical Society First edition published 2000 Second edition published 2003 Third edition published 2008 Fourth edition published 2012 Typeset by River Valley Technologies, India Printed in Great Britain by TJ International, Padstow, UK ISBN 978 0 85369 979 8 All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, without the prior written permission of the copyright holder. The publisher makes no representation, express or implied, with regard to the accuracy of the information contained in this book and cannot accept any legal responsibility or liability for any errors or omissions that may be made. The right of Donald Cairns to be identified as the author of this work has been asserted by him in accordance with the Copyright, Designs and Patents Act, 1988. A catalogue record for this book is available from the British Library.  Contents ix Preface to the fourth edition x Acknowledgements xi About the author 1 Chemistry of acids and bases 1 Dissociation of weak acids and bases 3 Hydrolysis of salts 9 Amphiprotic salts 10 Buffer solutions 11 Buffer capacity 13 Biological buffers 15 Ionisation of drugs 18 pK a values of drug molecules 19 pH indicators 19 Tutorial examples 22 Problems 26 2 Partition coefficient and biopharmacy 29 Experimental measurement of the partition coefficient 32 Drug absorption, distribution and bioavailability 35 Passive diffusion 38 The pH partition hypothesis 40 Active transport mechanisms 43 The action of local anaesthetics 44 Excretion and reabsorption of drugs 47 Food and drink 48 Tutorial examples 50 Problems 55 vi Contents 3 Physicochemical properties of drugs 57 Carboxylic acids 57 Phenols 61 Warfarin 62 Phenylbutazone 65 Indometacin 65 Barbiturates 65 Phenytoin 67 Sulfonamides 67 Basic drugs 69 Basicity of heterocyclic compounds 71 Separation of mixtures 71 Tutorial examples 73 Problems 77 4 Stereochemistry 81 Polarimetry 82 Biological systems 86 Fischer projections 87 Stereochemistry case study: thalidomide 94 Geometrical isomerism 97 Tutorial examples 99 Problems 101 5 Drug metabolism 103 Metabolic pathways 104 Cytochromes P450 105 Enzyme induction and inhibition 109 Drug conjugation reactions (Phase 2) 110 Stereochemistry 116 Metabolic pathways for common drugs 117 Tutorial example 125 Problems 126 6 Volumetric analysis of drugs 129 Volumetric flask 130 Pipette 130 Burettes 131 Units of concentration 131 Worked example 133 Concentration of active ingredients 134  Contents vii Design of an assay 134 Practical points 137 Back and blank titrations 140 Assay of unit-dose medicines 143 Non-aqueous titrations 144 REDOX titrations 145 Compleximetric titrations 148 Argentimetric titrations 149 Limit tests 150 Problems 151 7 Analytical spectroscopy 155 Effect of pH on spectra 159 Instrumentation 163 Experimental measurement of absorbance 166 Dilutions 167 Quantitative aspects of spectroscopy 168 Beer’s and Lambert’s laws 168 Methods of drug assay 170 Derivative spectroscopy 172 Infrared spectroscopy 174 Fluorimetry 177 Structure elucidation 178 Tutorial examples 188 Problems 196 8 Chromatographic methods of analysis 199 Definitions used in chromatography 199 Types of chromatography 200 Terms used in chromatography 211 Tutorial examples 214 Problems 216 9 Stability of drugs and medicines 217 Oxidation 217 Hydrolysis 228 Other mechanisms of degradation 233 Prodrugs 233 Tutorial examples 235 Problems 237 viii Contents 10 Kinetics of drug stability 239 Rate, order and molecularity 239 Rate equations and first-order reactions 240 Half-life 243 Shelf-life 243 Second-order reactions 243 Zero-order reactions 245 Reaction rates and temperature 245 Tutorial example 247 Problems 248 11 Licensing of drugs and the British Pharmacopoeia 249 European licensing procedures 250 Applications for marketing authorisations 252 British Pharmacopoeia Commission 253 The British Pharmacopoeia 254 12 Medicinal chemistry: the science of rational drug design 259 How do drugs work? 259 Where do drugs come from? 266 Why do we need new drugs? 271 13 Answers to problems 277 Selected bibliography 293 Index 295 Preface to the fourth edition In 2007 I wrote in the Preface to the third edition that the science and practice of pharmacy were undergoing a period of change. Now, as I write the Preface to the fourth edition of Essentials of Pharmaceutical Chemistry, it seems that the profession is still changing. The General Pharmaceutical Council has replaced the Royal Pharmaceutical Society as the regulator for pharmacy, the number of schools of pharmacy in the UK continues to rise and there are plans to integrate the current four year MPharm degree into a five year integrated model incorporating two six month periods in practice. This new edition of EPC (as one of my students calls the book) has been prepared against this changing background. I have tried to incorporate sug- gestions from users and reviewers and have updated every chapter for this edition and included new chapters, Chromatographic methods of analysis and Medicinal Chemistry: the Science of Rational Drug Design. As with previous editions, I am very grateful for the help, advice, comments and encouragement from colleagues, reviewers and the many students of pharmacy and pharmaceutical science who are forced to use this book as part of their studies. I hope they find this new edition useful and, perhaps, experience some of the wonder and enjoyment I felt when I was introduced to the chemistry of drugs and medicines more than 30 years ago. If I was being honest with myself, that’s still the way I feel about the subject today. Donald Cairns Aberdeen June 2011 Acknowledgements This book could not have been completed without the help of a great many people. I am very grateful to my colleagues, past and present for their advice and encouragement and particularly, for allowing me to assimilate their good practice (with or without their knowledge). This book would be poorer without their efforts. Special thanks must go to Paul Hambleton who read and commented on my first drafts and who not only allowed me to use a great many of his examination questions but also provided most of the answers! I am grateful to Louise McIndoe, Christina de Bono and all the staff at the Pharmaceutical Press for keeping me on track when diversions threat- ened and giving helpful advice about indexes, content pages, etc. Finally, I must thank my wife, Elaine, who looked after the weans while I bashed the keyboard upstairs. About the author Donald Cairns obtained a Bachelor of Science degree in pharmacy from the University of Strathclyde in 1980 and after a pre-registration year spent in hospital pharmacy, he returned to Strathclyde to undertake a PhD on the synthesis and properties of benzylimidazolines. Following a year as a post-doctoral research fellow in the department of pharmacy at Sunder- land Polytechnic (now the University of Sunderland), Dr Cairns moved to Leicester Polytechnic (now De Montfort University) where he held a five-year lectureship in pharmacy. In 1992 Dr Cairns was appointed senior lecturer in medicinal chemistry in Sunderland School of Pharmacy and in 2003 moved to a post of Associate Head of the School of Pharmacy at The Robert Gordon University in Aberdeen. In 2006, he was promoted to Professor of Pharmaceutical and Medicinal Chemistry at RGU and in 2010 was appointed Acting Head of the School of Pharmacy and Life Sciences. Professor Cairns has served as external examiner at Strathclyde, Liver- pool, Aberdeen and Belfast Schools of Pharmacy and has authored over 70 peer reviewed research papers. His research interests include the design and synthesis of selective anti- cancer agents, the molecular modelling of drug–DNA interactions and the design of prodrugs for the treatment of nephropathic cystinosis. Donald Cairns is a member of the General Pharmaceutical Council, the Royal Pharmaceutical Society, the Association of Pharmaceutical Scientists and in 2008 was made a Fellow of the Royal Society of Chemistry. In 2006 he was appointed to the British Pharmacopoeia Commission and serves on an Expert Advisory Group of the Commission on Human Medicines. To travel hopefully is a better thing than to arrive, and the true success is to labour. Robert Louis Stevenson, 1850–1894 1 Chemistry of acids and bases Chemistry is the defining science of pharmacy. To understand anything about a drug: the synthesis, the determination of its purity, the formulation into a medicine, the dose given, the absorption and distribution around the body, the molecular interaction of drug with its receptor, the metabolism of the drug and, finally, the elimination of drug from the body, requires a thorough and comprehensive understanding of the chemical structure of the drug and how this chemical structure influences the properties and behaviour of the drug in the body. For these reasons, chemistry is the most important of all the scientific disciplines contributing to the understanding of drugs and their actions in the body. A good understanding of the chemistry of drugs will allow the study of advanced topics such as drug design and medicinal chemistry, molecular pharmacology and novel drug delivery systems that are usually encountered in the later stages of a pharmacy or pharmaceutical science degree. Most of the drugs used in medicine are small organic molecules that be- have in solution as either weak acids or weak bases. In order to understand and appreciate these compounds a study must be made of simple acid–base theory. In 1887, the Swedish chemist Svante August Arrhenius suggested that solutions that conduct electricity (so-called electrolytes) do so because they dissociate into charged species called ions. Positively charged ions (or cations) migrate towards the negative terminal, or cathode, while negatively charged ions (or anions) migrate towards the positive terminal, or anode. It is this movement of ions that allows the passage of electric current through the solution. Compounds of this type may be classified as strong electrolytes, which dissociate almost completely into ions in solution, or as weak electrolytes, which only dissociate to a small extent in solution. Since strong electrolytes are almost completely dissociated in solution, measurement of the equilib- rium constant for their dissociation is very difficult. For weak electrolytes, however, the dissociation can be expressed by the law of mass action in terms of the equilibrium constant. 2 Essentials of Pharmaceutical Chemistry Considering the reaction A+B→C+D The equilibrium constant (K ) for the reaction is given by the product of the concentrations of the reaction products divided by the product of the concentrations of the reactants, or [C] × [D] K = [A] × [B] Clearly, if the equilibrium lies to the right-hand (or products) side, the numerator in the above expression will be greater than the denominator, and K will be greater than 1. Conversely, if the reaction does not proceed very far and the equilibrium lies closer to the left hand side, [A] × [B] will be larger than [C] × [D] and K will be less than 1. Strictly speaking, the law of mass action states that ‘the rate of a chemical reaction is proportional to the active masses of the reacting substances’, but for dilute solutions active mass may be replaced by concentration, which is much easier to measure. The law of mass action can be applied to the dissociation of water, a weak electrolyte widely used as a solvent in biological and pharmaceutical systems: H2 O H + + OH − The equilibrium constant for this reaction is given by [H + ] × [OH − ] K = [H2 O] In pure water, and in dilute aqueous solutions, the concentration of molecu- lar water, [H2 O], is so large as to be considered constant (approximately 55.5 M ), so the above expression simplifies to K w = [H + ] × [OH − ] (1.1) and K w is called the ionic product or autoprotolysis constant of water. The value of this equilibrium varies with temperature but is usually quoted as 1 × 10 −14 at 25 ◦ C. The units of K w are mole litre −1 × mole litre −1 , or mole2 litre −2 (also sometimes written as mole2 dm −6 , where 1 dm3 = 1 L). Since, in pure water, [H + ] = [OH − ], the hydrogen ion concentration in water is given by the square root of K w , which is 1 × 10 −7 mol L −1. Solutions in which the hydrogen ion concentration is greater than 10 −7 mol L −1 are called acidic, while solutions with a concentration of hydrogen ions of less than 10 −7 mol L −1 are referred to as alkaline. The range of hydrogen ion concentrations encountered in chemistry is very large, so it is convenient to adopt the pH notation first developed  Chemistry of acids and bases 3 by another Scandinavian chemist (Danish this time), Søren Peder Lauritz Sørensen. He defined pH as ‘the negative logarithm (to the base 10) of the hydrogen ion concentration’, or 1 pH = −log[H + ] = log [H + ] Use of the pH notation allows all degrees of acidity and alkalinity normally encountered in chemistry to be expressed on a scale from 0 to 14, corres- ponding to the concentrations of hydrogen ions contained in the solution. Solutions with a pH less than 7 are considered acidic, solutions with a pH greater than 7 are alkaline, while a solution with a pH of 7 is neutral. Sørensen carried out his important work on the development of the pH scale while working at the Carlsberg Laboratory in Copenhagen — something to think about next time you consume a bottle of his sponsor’s product! It should be noted that a sample of water will often give a pH reading of less than 7, particularly if the sample has been left in an open beaker. This is due to carbon dioxide present in the atmosphere dissolving in the water to give carbonic acid (H2 CO3 ), which dissociates to release hydrogen ions. Dissociation of weak acids and bases Acids are compounds that ionise to release hydrogen ions, or protons, to their surroundings. Bases are compounds that can accept hydrogen ions. This is called the Brønsted–Lowry definition of acids and bases (named after yet another Scandinavian chemist, Johannes Nicolaus Brønsted, and Thomas Martin Lowry, who was British). There are other ways of explaining acidity and basicity, but the Brønsted–Lowry theory works most of the time, and will be used throughout this book. The dissociation of a weak acid is usually represented as follows: HA H+ + A− However, this suggests that protons exist free in solution like little tennis balls bouncing around chemical reactions. The reality is that protons are solvated in solution: that is they go around attached to a solvent molecule. Since the most common solvent in pharmaceutical and biological systems is water, the ionisation of a weak acid is better represented as HA + H2 O H3 O + + A − where H3 O + is a hydroxonium ion, and the ionisation of a base can be represented as B + H2 O BH + + OH − 4 Essentials of Pharmaceutical Chemistry It is important to notice that water appears in these equations as both a proton acceptor and a proton donor. This is an example of the amphoteric (sometimes termed the amphiprotic) nature of water. Although the ionisa- tion of acids and bases in water is best described using the equations above, it is convenient to disregard the water when deriving useful expressions and relationships. Consider any weak acid HA, which dissociates as shown below: HA H+ + A− The equilibrium constant for this reaction is given, as before, by [H + ] × [A − ] K = [HA] In the case of an acid dissociation, the equilibrium constant for the reaction is termed K a , the ionisation constant, the dissociation constant or, some- times, the acidity constant. The above equation can now be rewritten as [H + ] × [A − ] Ka = [HA] For exact work, the concentration term must be replaced by the thermo- dynamic activity of the ion, but for dilute solutions concentration may be used. For a given compound at a given temperature, K a is a constant. Clearly, the farther the above equilibrium lies to the right-hand side, the more completely the acid will ionise and the greater will be the value of K a. To put it more simply, the greater the value of K a , the stronger is the acid. Using the equation above, it is possible to derive an expression for the strength of acid solutions. If the acid HA ionises to α mol hydrogen ions and α mol hydroxyl ions, where α is the fraction of the acid that is ionised, then the number of moles of undissociated acid is given by (1 − α ). This acid solution can now be prepared with c mol of acid in 1 litre (or 1 dm3 ), which will yield α c mol hydrogen ions and α c mol A −. Hence, HA H+ + A− (1 − α ) c αc + αc αc × αc Ka = (1 − α )c α 2 c2 Ka = (1 − α )c α2c Ka = (1 − α )  Chemistry of acids and bases 5 For weak electrolytes, α is very small and may be neglected so (1 − α ) is approximately equal to 1. The simplified expression may now be written as Ka = α 2 c where c is the concentration, in moles per litre, and α is the degree of ionisation of the acid. Then s  Ka α = c The pH of the solution can now be determined: [H + ] = α c Therefore, s  Ka p + [H ] = c = (K a c) c Taking logarithms, log[H + ] = 1 2 log K a + 1 2 log c Multiplying throughout by −1 gives − log[H + ] = − 21 log K a − 1 2 log c Therefore, 1 1 pH = 2 pK a − 2 log c (1.2) Equation (1.2) applies to the ionisation of weak acids, but a similar expres- sion can be derived for weak bases. The equation for the ionisation of a weak base may be expressed as B + H2 O BH + + OH − (1 − α )c αc + αc where B is the base and BH + is termed the conjugate acid of the base. The equilibrium constant for this reaction is written as [BH + ] × [OH − ] Kb = [B] where K b is termed the base dissociation constant or basicity constant. α2c Kb = (1 − α ) 6 Essentials of Pharmaceutical Chemistry As before α is very small and can be neglected, so (1 − α ) is approximately equal to 1. α 2 = K b /c α = ( K b / c) p From above, [OH − ] = cα Therefore, p (K b / c ) = (K b c ) p [OH − ] = c However, Kw [OH − ] = [H + ] Therefore, Kw p + = ( Kb c ) [H ] and Kw [H + ] = √ ( Kb c ) Taking logarithms, log[H + ] = log K w − 1 2 log K b − 1 2 log c or 1 1 pH = pK w − 2 pK b + 2 log c (1.3) Eqs. (1.2) and (1.3) are extremely useful because they allow the pH of solutions of weak acids and bases to be calculated if the concentrations and dissociation constant are known. How strong an acid is depends on how many hydrogen ions are released when the acid ionises, and this depends on the degree of ionisation, α, for any given concentration. As stated above, K a , the equilibrium constant for the dissociation of the acid, gives a measure of how far the ionisation equilibrium lies to the right-hand, or products, side. As can be seen from Equation (1.3), the similar expression, K b , gives a measure of basic strength and, as with K a , the higher the numerical value of K b , the stronger is the base. It is often useful and convenient to express the strengths of acids and bases using the same term, pK a , and this can be done by considering the  Chemistry of acids and bases 7 equilibria that exist between an acid and its conjugate base. A weak acid (HA) and its conjugate base (A − ) are related as follows: HA H+ + A− A − + H2 O HA + OH − From the equations above, [H + ] × [A − ] Ka = [HA] and [HA] × [OH − ] Kb = [A − ] Then [H + ] × [A − ] [HA] × [OH − ] Ka × Kb = × [HA] [A − ] Cancelling similar terms gives K a × K b = [H + ] × [OH − ] which can be rewritten as K a × K b = K w = 1 × 10 −14 (1.4) That is, the acid dissociation constant and the base dissociation constant are related through the ionic product of water. Equation (1.4) is a very important relationship since it allows the calcu- lation of K b or K a if the other is known. It also follows that the strengths of acids and their conjugate bases are related through K w. This means that a strong acid must have a weak conjugate base and, similarly, a weak acid must have a strong conjugate base. A moment’s thought will confirm that this must, indeed, be true. Acids and their conjugate bases are related by equilibria, which can be thought of as giant seesaws. If one partner of the pair is very strong and heavy, the other will be weak and light. The same relationship applies to acid–conjugate base equilibria. This relationship also allows chemists to be lazy and express the strengths of acids and bases in terms of the dissociation constant for the acid. This is particularly true when we consider the term pK a. In a similar manner to pH, the pK a of an acid is defined as the negative logarithm (to the base 10) of the dissociation constant, K a , i.e. pK a = − log K a This terminology allows chemists to talk loosely about the pK a of acids and bases, when what they really mean is the pK a of acids and the conjugate 8 Essentials of Pharmaceutical Chemistry acids of bases. It is incorrect to say ‘the pK a of a primary amine is between 9 and 10’, although the usage is widespread. It is more accurate to say ‘the pK a of the conjugate acid of a primary amine is between 9 and 10’. This is just another example of lecturers saying one thing and meaning another. Another source of confusion concerning strengths of acids arises with K a and pK a. The term K a is the dissociation constant for the ionisation of an acid, and hence the larger the value of K a , the stronger is the acid (since the equilibrium constant lies farther to the right-hand side). pK a is the negative logarithm of K a and is used commonly because K a values for organic acids are very small and hard to remember (typically 10 −5 ). It follows that since pK a is the negative logarithm of K a , the smaller the value of pK a the stronger is the acid. Consider the two carboxylic acids below: Acetic acid, CH3 COOH, pK a = 4.7 Chloroacetic acid, ClCH2 COOH, pK a = 2.7 In answer to the question, ‘which acid is the stronger?’ clearly it is chloroacetic acid, since its pK a is smaller. A student of organic chemistry could even suggest that the reason is due to increased stabilisation of the anion formed on ionisation by the electronegative chlorine atom. If the question is asked ‘how much stronger is chloroacetic than acetic?’, then all sorts of interesting answers appear, ranging from ‘twice as strong’ to a ‘million times as strong’. The answer, obvious to anyone who is familiar with logarithms, is that chloroacetic is 100 times stronger than acetic acid. This is because the difference in pK a is two units on a log scale, and the antilog of 2 to the base 10 is 100. It is important for students (and graduates!) to appreciate that pH and pK a are logarithmic relationships and that a K value corresponding to a pK a of 2.7 is not really close to a K value corresponding to a pK a of 4.7. Equation (1.4) can be rewritten in a logarithmic form by taking the negative logarithm of both sides, to give pK a + pK b = pK w = 14 (1.5) In addition, since pK b may be rewritten as pK w − pK a this allows Eq. (1.3) to be rewritten omitting any reference to pK b : 1 1 1 pH = 2 pK w + 2 pK a + 2 log c or 2 (pK w 1 pH = + pK a + log c)  Chemistry of acids and bases 9 Hydrolysis of salts When a salt is dissolved in water, the compound dissociates completely to give solvated anions and cations. This breaking of bonds by the action of water is called hydrolysis and the salt is said to be hydrolysed. The pH of the resulting solution depends on whether the salt was formed from reaction of strong or weak acids and bases and there are four possible combinations. For example, if the salt results from reaction between a strong acid and a strong base (e.g. NaCl), then the resulting solution will be neutral, and NaCl is termed a neutral salt. Of the two ions produced, Na + and Cl − , only the Cl − reacts with water: Cl − + H2 O HCl + OH − This reaction does not occur to any great extent since the Cl − is the conjugate base of a strong acid, namely HCl. The choride ion is, therefore, a very weak conjugate base and its reaction with water can be neglected. If the salt results from reaction between a strong acid and a weak base (e.g. the reaction of ammonia and hydrogen chloride to give ammonium chloride) HCl + NH3 NH4+ + Cl − then the resulting salt solution will be acidic by hydrolysis and the pH of an aqueous solution of the salt will be less than 7. This can be demonstrated by considering the reactions that occur when ammonium chloride is hydrolysed. The salt dissociates completely to give hydrated ammonium ions and hydrated chloride ions. The chloride ion is not very reactive towards water, but the ammonium ions react with water to give ammonium hydroxide. This is because the ammonium ion, NH4+ , is the conjugate acid of the weak base NH3 and must, therefore, be quite strong. The ammonium ion reacts with water as follows to produce H3 O + : NH4+ Cl − NH4+ + Cl − NH4+ + H2 O NH3 + H3 O + An increase in the concentration of H3 O + results in a fall in pH, and an acidic solution. The pH of this solution can be calculated by using the equation derived for a weak acid, Eq. (1.2) above: 1 1 pH = 2 pK a − 2 log c If the salt results from the reaction of a strong base and weak acid (e.g. sodium acetate from reaction of sodium hydroxide and acetic acid), then 10 Essentials of Pharmaceutical Chemistry the solution formed on hydrolysis will be basic, i.e. NaOH + CH3 COOH CH3 COO − Na + + H2 O CH3 COO − Na + + H2 O CH3 COOH + OH − + Na + Na + does not react with water to any great extent, but CH3 COO − is the conjugate base of the weak acid CH3 COOH and is, therefore, strong enough to react with water to produce OH −. The increase in concentration of OH − gives a basic solution, the pH of which can be calculated from the equation for the pH of weak bases, Eq. (1.3). 1 1 pH = pK w − 2 pK b + 2 log c or, if K b is replaced by the expression K w − K a 1 1 1 pH = 2 pK w + 2 pK a + 2 log c which is probably the easiest form to remember. The final scenario involves a salt formed between a weak acid and weak base (e.g. ammonium acetate, NH4+ CH3 COO − ). The hydrogen and hydroxyl ions formed by hydrolysis of ammonium acetate occur in roughly equal concentrations, which will yield a neutral salt. These relationships can be summarised as follows: S t ro n g a c i d + S t ro n g b a s e → N e u t ra l s a l t S t ro n g a c i d + We a k b a s e → A c i d i c s a l t We a k a c i d + S t ro n g b a s e → B a s i c s a l t We a k a c i d + We a k b a s e → N e u t ra l s a l t This does seem to follow a type of logic. Using the seesaw analogy for equilibria again, if both partners are strong or both are weak, then the seesaw balances, and the solution formed by hydrolysis is neutral. If either partner is strong, then the seesaw tilts to that side to give an acidic or basic solution. This analogy is not precise, but it may help the desperate student remember the pH values of hydrolysed salt solutions. Amphiprotic salts The reactions of salts in water become more complicated if the salt in question is amphiprotic; that is, it can function both as an acid and a base. Examples of amphiprotic anions are bicarbonate (sometimes called hy- drogencarbonate), HCO3− , and bisulfite (or hydrogensulfite), HSO3−. These species can donate or accept hydrogen ions in solution.  Chemistry of acids and bases 11 The pH of a solution of an amphiprotic salt (e.g. sodium bicarbonate, Na + HCO3− ) is given by the equation 2 (pK a1 + pK a2 ) 1 pH = (1.6) where pK a1 and pK a2 refer to the ionisation constants for the acid and base reactions, respectively. In the case of sodium bicarbonate, these values are 6.37 and 10.25, which means the pH of any concentration of sodium bicarbonate will be 8.31 and the solution will be slightly basic. Buffer solutions A buffer solution is a solution that resists changes in pH. If acid is added then, within reason, the pH does not fall; if base is added, the pH does not rise. Buffers are usually composed of a mixture of weak acids or weak bases and their salts and function best at a pH equal to the pK a of the acid or base involved in the buffer. The equation that predicts the behaviour of buffers is known as the Henderson–Hasselbalch equation (named after chemists Lawrence Joseph Henderson and Karl Albert Hasselbalch), and is another vitally important equation worth committing to memory. It is derived as follows, by considering a weak acid that ionises in solution: HA H+ + A− The equilibrium constant for this ionisation is given by [H + ] × [A − ] Ka = [HA] Taking logarithms of both sides and separating the hydrogen ion term gives [A − ] log K a = log[H + ] + log [HA] Multiplication throughout by −1 gives [A − ] − log K a = − log[H + ] − log [HA] or [A − ] pK a = pH − log [HA] which rearranges to give [A − ] pH = pK a + log [HA] Since the acid in question is weak, the number of ions of the A − derived from dissociation of the acid itself is very small compared with 12 Essentials of Pharmaceutical Chemistry the number derived from the fully ionised salt. This means that [A − ] is approximately equal to total concentration [SALT]; similarly [HA], since the acid is weak and predominantly unionised, is approximately equal to the total acid concentration [ACID]. The equation can now be rewritten as [SALT] pH = pK a + log (1.7) [ACID] The Henderson–Hasselbalch equation can also be derived from considera- tion of the ionisation of a weak base, B, which ionises in aqueous solution as follows: B + H2 O BH + + OH − In this case the [SALT] term can be replaced by the concentration of the conjugate acid of the weak base, [BH + ], which, in effect, yields the same equation. An example of a buffer is a mixture of acetic acid and sodium acetate, which will ionise as follows: CH3 COOH CH3 COO − + H + CH3 COO − Na + CH3 COO − + Na + Since the acetic acid only ionises to a small extent, there will be a high concentration of undissociated acid (shown in bold) or, to put it another way, the equilibrium for the reaction will lie predominantly to the left-hand side. Sodium acetate is a salt and will ionise completely to give high con- centrations of CH3 COO − and Na + (shown in bold). If hydrogen ions are now added to the buffer solution, they will react with the high concentration of CH3 COO − present to give undissociated acetic acid. Acetic acid is a weak acid and only dissociates to a small extent, so the pH of the solution does not decrease. In effect, the hydrogen ions of a strong acid are mopped up by the buffer to produce a weak acid, acetic acid, which is not sufficiently acidic to lower the pH. H + + CH3 COO − CH3 COOH Similarly, if hydroxyl ions are added to the buffer system, they will react with the high concentration of free acetic acid present to give water and acetate ions: OH − + CH3 COOH H2 O + CH3 COO − Neither water nor acetate is sufficiently basic to make the solution alkaline, so the pH of the buffer solution will not increase. The high concentration of sodium ions has little or no effect on the pH of the solution since when these ions react with water they do so to produce  Chemistry of acids and bases 13 equal numbers of hydrogen and hydroxyl ions as shown below: Na + + H2 O Na + OH − + H + and Na + OH − Na + + OH − Buffers can also be composed of weak bases and their salts; examples include ammonia buffer, used to control the pH of compleximetric titrations (see Chapter 6) and the common biological buffer TRIS (or tris(hydroxymethylaminomethane), C4 H11 NO3 ), used to control the pH of protein solutions. Buffer capacity Buffer solutions work best at controlling pH at pH values roughly equal to the pK a of the component acid or base: that is, when the [SALT] is equal to the [ACID]. This can be shown by calculating the ability of the buffer to resist changes in pH, which is the buffer capacity. The buffer capacity is defined as the number of moles per litre of strong monobasic acid or base required to produce an increase or decrease of one pH unit in the solution. When the concentrations of salt and acid are equal, the log term in the Henderson–Hasselbalch equation becomes the logarithm of 1, which equals 0. To move the pH of the buffer solution by one unit of pH will require the Henderson–Hasselbalch equation to become 10 pH = pK a + log 1 It will require addition of more acid or base to move the pH by one unit from the point where pH = pK a than at any other given value of the ratio. This can be neatly illustrated by the following example. Suppose 1 litre of buffer consists of 0.1 M CH3 COOH and 0.1 M CH3 COO − Na + : the pH of this buffer solution will be 4.7 (since the log term in the Henderson–Hasselbalch equation cancels). Now, if 10 mL of 1 M NaOH is added to this buffer, what will be the new pH? Clearly, the 10 mL of NaOH will ionise completely (strong alkali) and some of the 0.1 M acetic acid will have to convert to acetate anion to compensate. The new pH will be [SALT] pH = pK a + log [ACID] (0.1 + 0.01) pH = 4.7 + log (0.1 − 0.01) 0.11 pH = 4.7 + log 0.09 pH = 4.79 14 Essentials of Pharmaceutical Chemistry The addition of 10 mL of 1 M alkali has only increased the pH of the buffer by a small amount. By way of comparison, if 10 mL of 1 M NaOH were added to 1 litre of pure water, the pH of the solution would increase from a pH of 7 to a value of approximately 12. This can be easily shown by using the term pOH, which is defined as the negative logarithm of the hydroxyl ion concentration in a similar way to pH = − log[H + ]. The term pOH is used much less frequently in the literature than pH but it follows that if pOH for 0.01 M NaOH = 2 and pH + pOH = 14, the pH of the solution in the example above is 12. The buffer capacity (β ) for this buffer can now be calculated as No. of moles of NaOH added β = Change in pH observed 0.01 β = (4.79 − 4.7) 0.01 β = 0.09 β = 0.11 Since buffer solutions work best at a pH equal to the pK a of the acid or base of which they are composed, consideration of the pK a will determine choice of buffer for a given situation. The pK a of acetic acid is 4.7, and therefore an acetic acid–acetate buffer would be useful for buffering a solution to a pH of approximately 5. Similarly, an alkaline buffer can be obtained by using ammonia solution, which will buffer to a pH of approximately 10 (pK a of ammonia is 9.25). If a buffer is required to control the pH of a neutral solution, use is made of the second ionisation of phosphoric acid. Phosphoric acid is a triprotic acid, which requires three equivalents of NaOH as follows: H3 PO4 + NaOH Na + H2 PO4− + H2 O pK a = 2.12 Na H2 PO4− + NaOH + (Na ) 2 HPO2− + 4 + H2 O pK a = 7.21 − (Na + ) 2 HPO24 + NaOH (Na + ) 3 PO3− 4 + H2 O pK a = 12.67 A mixture of sodium dihydrogen phosphate, Na + H2 PO4− , and disodium hydrogen phosphate, (Na + ) 2 HPO2− 4 , will function as a buffer and control the pH to a value of approximately 7.0. In this example, the species with the greater number of available hydrogen atoms functions as the acid (i.e. Na + H2 PO4− ), while the (Na + ) 2 HPO2− 4 functions as the salt.  Chemistry of acids and bases 15 The choice of buffer to use in a given situation, therefore, depends on the pK a of the acid or base involved. As a general rule, buffer solutions work well within + 1 or − 1 pH unit of the pK a. Beyond these values, the buffer capacity is too small to allow effective buffer action. Biological buffers The human body contains many buffer systems, which control the pH of body compartments and fluids very effectively. Blood plasma is maintained at a pH of 7.4 by the action of three main buffer systems: first, dissolved carbon dioxide, which gives carbonic acid (H2 CO3 ) in solution, and its sodium salt (usually sodium bicarbonate, NaHCO3 ). This is responsible for most of the buffering capacity. The other two buffers are dihydrogen phosphate (H2 PO4− ), also with its sodium salt, and protein macromolecules. Proteins are polymers composed of repeating units called amino acids. These amino acids (as their name suggests) are compounds containing NH2 and COOH groups in the same molecule and have the general formula shown in Fig. 1.1. COOH H C R NH2 Figure 1.1. The general formula of amino acids. Proteins are composed of about 20 different amino acid residues, which are connected to each other by peptide bonds formed between one amino acid and its neighbour. The side-chain of the amino acid may be acidic (as in the case of glutamic and aspartic acids), basic (as in the case of arginine and lysine) or neutral (as in alanine). A protein, which may be composed of hundreds of amino acid residues, is, therefore, a polyelec- trolyte whose properties depend on the balance of acidic and basic groups on the side-chains. Generally, most proteins act as weak acids and form buffers with their sodium salts. Compounds such as amino acids, which are capable of acting as both acids and bases, are known as amphoteric, or sometimes, amphiprotic. In solution, free amino acids usually do not exist in the molecular form shown in Fig. 1.1, but instead both the amino and carboxyl groups ionise to form an internal salt, as shown in Fig. 1.2. These internal salts are known by the German word zwitterion (‘dipolar ion’), and formation of the zwitterion makes the amino acid very polar and, therefore, very soluble in water. If acid is added to the zwitterion, the ionised COO − group will accept a proton to give undissociated COOH. 16 Essentials of Pharmaceutical Chemistry COO– H C + R NH3 Figure 1.2. The structure of a zwitterion. The overall charge on the amino acid will now be positive, due to the NH3+. Similarly, if base is added to the zwitterion, the NH3+ (which is really the conjugate acid of NH2 ) will function as an acid and donate its proton to the base. The overall charge on the amino acid will now be negative, due to the ionised COO −. Amino acids are, therefore, ionised at all values of pH. They are positively charged at low pH, negatively charged at high pH and zwitterionic at neutral pH. The fact that amino acids are ionised at all values of pH and are zwitterionic at neutral pH has profound implications for the oral absorption and bioavailability of amino acids from the diet. The body has to resort to specialised uptake mechanisms to ensure that sufficient levels of these essential nutrients are absorbed (see Chapter 2). The ionisation of the simplest amino acid, glycine, is represented in Fig. 1.3. COOH + COO– + COO– –H –H H H H C + C + C H NH3 H NH3 H NH2 Figure 1.3. The ionisation of glycine. If the pH of the protein or amino acid solution is adjusted so that the number of ionised COO − groups is equal to the number of ionised NH3+ groups, then that value of pH equals pI , the isoelectric point of the protein or amino acid. This point corresponds to the minimum solubility of the protein, and the point at which migration of the protein in an electric field is slowest (as in the technique of electrophoresis, which is used to separate mixtures of proteins according to their overall electrical charge). The isoelectric point for an amino acid may be easily calculated if the pK a values for the NH3+ and COO − are known (e.g. by titration). For a simple amino acid, such as glycine, the pI is simply the average of the two pK a values. For more complex amino acids, such as glutamic acid or arginine, which have ionisable groups in the side-chains, the pI is given by averaging the two pK a values that lie on either side of the zwitterion. This is true no matter how many times an amino acid or peptide ionises. For an amino acid with one acidic group on the sidechain, there are three distinct ionisations  Chemistry of acids and bases 17 and hence three distinct pK a values. Fully protonated aspartic acid ionises as shown in Fig. 1.4. COOH COO – H pKa1 = 1.88 H HOOC C + HOOC C + + H+ NH3 N H3 COO – COO – H pKa2 = 3.65 H HOOC C + –OOC C + + H+ NH3 N H3 COO – COO – H pKa3 = 9.60 H –OOC C + –OOC C + H+ NH3 NH2 COO – Na+ H C HOOC NH2 MSG Figure 1.4. The ionisation of aspartic acid and structure of MSG. The first group to ionise (and hence the strongest acid) is the COOH group on the α-carbon. This gives pK a1. The second proton is lost from the side-chain COOH to give pK a2. Finally, the NH3+ on the α-carbon ionises to give pK a3. There is, of course, only one pI , which is given by the average of the two pK a values on either side of the zwitterion: 12 (pK a1 + pK a2 ). The other commonly occurring amino acid with an acidic side-chain is glutamic acid. This compound is probably best known as its monosodium salt (monosodium glutamate (MSG)). This salt is added to foods (especially oriental food) to enhance the flavour and impart a ‘meat-like’ taste to the food. Interestingly, both the D -enantiomer of glutamic acid and the nat- urally occurring L -form are used as food additives. Use of the non-natural D -isomer may account for some of the adverse reactions experienced by consumers of MSG in food. 18 Essentials of Pharmaceutical Chemistry Ionisation of drugs When a weakly acidic or basic drug is administered to the body, the drug will ionise to a greater or lesser extent depending on its pK a and the pH of the body fluid in which it is dissolved. The pH of the body varies widely, but the most important biological solution is the blood, which, as stated above, normally has a pH of 7.4. An equation can be derived that will predict the extent to which the drug ionises, and, as is often the case, the starting point for the derivation is the Henderson–Hasselbalch equation (Eq. (1.7)). [SALT] pH = pK a + log [ACID] [A − ] pH = pK a + log [HA] Rearranging, [HA] pK a − pH = log [A − ] and, therefore, [HA] = [A − ] × antilog(pK a − pH) The fraction of the total drug that is ionised is given by [A − ] [HA] + [A − ] so that the fraction ionised is [A − ] [A − ] × antilog(pK a − pH) + [A − ] which simplifies to 1 Fraction ionised = (1.8) 1 + antilog(pK a − pH) Equation (1.8) applies to drugs that are weak acids and allows the fraction of the total dose that is ionised to be calculated for any pH if the pK a of the drug is known. The equation is sometimes written as the percentage ionised, which is simply given by 100 % Ionised = (1.9) 1 + antilog(pK a − pH)  Chemistry of acids and bases 19 A similar expression can be derived for drugs that are weak bases, to give Eqs. (1.10) and (1.11) below. 1 Fraction ionised for basic drug = (1.10) 1 + antilog(pH − pK a ) and 100 % Ionised for basic drug = (1.11) 1 + antilog(pH − pK a ) pKa values of drug molecules Most compounds used in medicine are either weak acids or weak bases (and quite a few are both!). This means that the range of possible pK a values encountered in drug molecules is huge. It is important to remember that the value of the pK a for a drug tells you absolutely nothing about whether the compound is an acid or base. The pK a value is simply the negative logarithm of the dissociation constant and can, within reason, have any value. This contrasts with the pH notation, where a pH value 7 means that it is alkaline. It would be quite wrong to say that because one particular acid has a pK a of 3, then all compounds with a pK a of 3 must be acids. Many weak bases have pK a values of 2 to 4. Similarly, while a basic drug like cocaine has a pK a of 9.5, this does not mean that all compounds with a pK a greater than 7 are bases. Indeed, phenols, which are weak acids, mostly have pK a value of approximately 10. Only a thorough understanding of chemical structure and a knowledge of the functional groups that confer acidity or basicity on a molecule will allow the correct prediction of the acidic or basic nature of a molecule. To illustrate this, Table 1.1 lists some common acidic and basic drugs with their pK a values. pH indicators In Chapter 6, the long-suffering reader will encounter volumetric analyses. This technique involves the accurate addition of volumes of solution in order to determine the purity of drugs and raw materials. The end point of many of these titrations can be determined by the colour change of an indicator. The indicators used in pH titrations are themselves weak acids or bases that change colour depending on whether they are ionised or not. The best indicators change colour sharply at a given pH, and tables of indicators and their pH range are available. The ionisation of indicators is determined by the Henderson–Hasselbalch equation, where pK a refers to 20 Essentials of Pharmaceutical Chemistry Table 1.1 pKa values of some common drugs Drug pKa value Acidic drugs Aspirin 3.5 Paracetamol 9.5 Phenobarbital 7.4 (first ionisation) Basic drugs Cocaine 8.6 Diazepam 3.3 Diphenhydramine 9.0 Amphoteric drugs Morphine 8.0 (amine), 9.9 (phenol) Adrenaline (epinephrine) 8.7 (amine), 10.2, 12.0 (phenols) the negative logarithm of the acid dissociation constant of the indicator, and [SALT] and [ACID] refer to the concentrations of the ionised and unionised forms of the indicator, respectively. If the indicator is a weak base, the Henderson–Hasselbalch equation has to be rewritten as [BASE] pH = pK a + log [ACID] since the salt term is really the conjugate acid of the weak base. The choice of an indicator for a titration can be made by predicting the pH at the end point of the titration. This is done accurately by working out the proportion of each species at the end of the titration, using the equations above, and determining also the pH due to hydrolysis of any salts present; it may be estimated (and a lot of miserable algebra avoided) as follows. If the pH of the end point solution is equal to the pK a of the acid or conjugate acid involved, then there will be equal concentrations of the ionised and unionised forms of the compound present. This is because if pH = pK a then the log term in the Henderson–Hasselbalch equation is 1 and [unionised] = [ionised]. If the pH of the solution is increased to one unit above the pK a of the acid (or one unit below the pK a of the conjugate acid), then the percentage of the compound ionised increases to about 90%. If the pH increases to two units above the pK a (or two units below for a base), the percentage ionised increases to 99%, since both pH and pK a are logarithmic relationships, and so on to 99.9%, 99.99%, etc. This approximate ‘rule of thumb’ is summarised below.  Chemistry of acids and bases 21 For weak acids: pH = pK a compound is approximately 50% ionised pH = pK a + 1 compound is approximately 90% ionised pH = pK a + 2 compound is approximately 99% ionised pH = pK a + 3 compound is approximately 99.9% ionised pH = pK a + 4 compound is approximately 99.99% ionised For weak bases: pH = pK a compound is approximately 50% ionised pH = pK a − 1 compound is approximately 90% ionised pH = pK a − 2 compound is approximately 99% ionised pH = pK a − 3 compound is approximately 99.9% ionised pH = pK a − 4 compound is approximately 99.99% ionised This relationship is hugely important and well worth committing to memory. It will reappear many times in this book, in many different guises, and will allow the reader to impress colleagues (particularly medical colleagues) with their uncanny understanding of pH and ionisation of drugs. In the case of predicting the pH at the end point of titrations, most acid–base reactions are considered over when the ratio of ionised form to unionised form is 1000 to 1, i.e. when (99.9) pH = pK a + log (0.1) From the rules above, this point is reached when the pH of the solution is three units above the pK a of the acid (or three units below the pK a of the conjugate acid of the base), and this allows an appropriate indicator to be chosen. For example, if the acid being titrated has a pK a of 4.7, then the end point pH will be 4.7 + 3 = 7.7, and an indicator that changes colour between pH 7.0 and 8.0 should be chosen. Similarly, for a base with a pK a of 8.5, the end point pH will be 8.5 − 3 = 5.5, and an indicator with a pH range of 5.0–6.0 should be used. The pH ranges of many common indicators are shown in Chapter 6 (p. 140). 22 Essentials of Pharmaceutical Chemistry Tutorial examples Q 1 Ephedrine is a naturally occurring drug useful in the treatment of asthma. Its structure is shown in Fig. 1.5. (a) Classify ephedrine as acidic, basic or neutral. (b) Using your answer to part (a) as a guide suggest a simple way in which the water solubility of the drug could be increased. H OH CH 3 H N CH 3 H Figure 1.5. The structure of ephedrine, pKa = 9.6. A 1(a) Ephedrine is an alkaloid produced by Ephedra (the Ma huang plant). It was widely used for the relief of bronchospasm asso- ciated with an attack of asthma. The drug has been superseded in recent years by safer, more effective bronchodilators such as salbutamol and terbutaline. The diastereoisomer of ephedrine, pseudoephedrine, is widely used in cough mixtures as a decongest- ant. Ephedrine is a secondary amine and, because the lone pair of electrons on the nitrogen can react with hydrogen ions, is basic in solution (Fig. 1.6). H OH H OH CH 3 CH 3 + H2O + OH – + H N CH 3 H N CH 3 H H H Figure 1.6. Reaction of ephedrine with water. (b) The water solubility of the drug could be increased by forming a salt with a mineral acid such as hydrochloric acid to give ephedrine hydrochloride (Fig. 1.7).  Chemistry of acids and bases 23 H OH H OH CH 3 CH 3 + HCl + + Cl – H N CH 3 H N CH 3 H H H Figure 1.7. Reaction of ephedrine with hydrochloric acid. This salt will be acidic by partial hydrolysis (salt of a weak base and a strong acid). The pH of the salt solution is given by Eq. (1.2). 1 1 pH = 2 pK a − 2 log c If 1 M HCl is reacted with 1 M ephedrine, the resulting concen- tration of ephedrine hydrochloride will be 0.5 M ; therefore, 1 1 pH = 2 pK a − 2 log c 2 (9.6) − 2 log(0.5) 1 1 = = 4.8 − (−0.15) = 4.95 which, as predicted, is on the acidic side of neutral. Incidentally, note that the concentration of ephedrine hydrochloride formed above is not 1 M , which may be supposed initially. One mole of ephedrine does give one mole of salt, but the volume of the solution will double when the HCl is added, so the concentration will be halved. Q 2 Calculate the pH of 0.05 M sodium acetate, given that the pK a of acetic acid is 4.66. A 2 Since sodium acetate is the salt of a strong base and a weak acid, it will be basic by partial hydrolysis. We can, therefore, use Eq. (1.3) for weak bases to calculate the answer. pH = pK w − 12 (pK b − log c ) pH = 14 − 12 (14 − 4.66) + 1 2 log 0.05 pH = 14 − 4.67 + (−0.65) pH = 8.68 24 Essentials of Pharmaceutical Chemistry Q 3 Calculate the concentration of acetic acid to be added to a 0.1 M solution of sodium acetate to give a buffer of pH 5 (pK a of acetic acid is 4.66). A 3 Acetic acid is a weak acid, so its degree of ionisation is very small and the contribution to the total concentration of acetate anions from ionisation of the acid can be ignored. The total salt concentration is, therefore, 0.1 M from the fully ionised sodium acetate. Using the Henderson–Hasselbalch equation (Eq. (1.7)), [SALT] pH = pK a + log [ACID] 0.1 5.0 = 4.66 + log [ACID] 0.1 0.34 = log [ACID] 0.1 2.188 = [ACID] [ACID] = 0.046 M Q 4 Weak acids and bases are often formulated as their salts to make them more water soluble. The ionised salts, however, do not cross biological membranes very well. Calculate the percentage of a dose of pentobarbital that will be ionised at plasma pH (7.4). The structure of pentobarbital is shown in Fig. 1.8. O H H5 C 2 N O H11 C 5 N O H Figure 1.8. The structure of pentobarbital, pKa = 8.0.  Chemistry of acids and bases 25 A 4 Pentobarbital is a barbiturate and is a weak acid. Normally, compounds that contain a nitrogen atom are basic (ammonia, amines, some heterocycles, etc.), but these compounds are only basic if the lone pair of electrons on the nitrogen is available for reaction with hydrogen ions to form salts. In the case of pentobar- bital (and other barbiturates such as phenobarbital, butobarbital, etc.), the lone pair on the ring nitrogens is unavailable for reaction due to resonance with the adjacent carbonyl groups. Instead, the hydrogen on the nitrogen can be lost as a proton, and the resulting negative charge delocalised around the molecule, as shown in Fig. 1.9. This resonance-stabilised anion allows barbiturates to function as weak acids, and sodium salts may be formed to increase the water solubility of the drug and allow parenteral administration. O H O H5 C 2 N + H5C 2 N (–) –H O O H11 C 5 N H11 C 5 N O H O H O (–) H5 C 2 N O H11 C 5 N O H O H5 C 2 N O (–) H11 C 5 N O H O (–) H5 C 2 N O H11 C 5 N O H Figure 1.9. Resonance forms of the pentobarbital anion. 26 Essentials of Pharmaceutical Chemistry To calculate the percentage ionised, use can be made of equations of the type 100 % Ionised = 1 + antilog(pK a − pH) which is easily derived from the Henderson–Hasselbalch equation and will work for weak acids if the pK a is known. However, in this case an expression can easily be derived from first principles. If we let x be the % ionised, [SALT] pH = pK a + log [ACID] x 7.4 = 8.0 + log (100 − x ) x −0.6 = log (100 − x ) x = antilog(−0.6) = 0.251 (100 − x ) x = 0.251(100 − x ) x = 25.1 − 0.251x 25.1 x= = 20.1 1.251 % Ionised at pH 7.4 = 20.1% % Unionised at pH 7.4 = 79.9% Problems Q1.1 (a) Ethanolamine (HOCH2 CH2 NH2 , relative molecular mass 61.08) has a pK a of 9.4. Explain what this term means. (b) Explain why ethanolamine is freely soluble in water, and why the resulting solution is basic. (c) Calculate the pH of a 1% w/v solution of ethanolamine. (d) A solution of pH 9.0 is required that will resist changes in pH on the addition of small amounts of strong acid or  Chemistry of acids and bases 27 strong base. Indicate briefly a possible composition of such a solution, and show how pH changes are resisted. Q1.2 (a) What do you understand by the term pK a ? Explain how this value can be used to indicate the strength of a base. (b) The base ephedrine has a pK a value of 9.6. Calculate the the- oretical end point pH when a 0.1 M solution of ephedrine is titrated with 0.1 M HCl. (c) Acetic acid (CH3 COOH) has a pK a value of 4.76. How might you prepare an acetate buffer with a pH of 5.0, containing 0.1 mol L −1 of the acid? (d) Calculate the buffer capacity of the solution described above. Q1.3 Describe the ionisation or ionisations that occur when fully pro- tonated lysine (Fig. 1.10) is subjected to increasing pH. What is the dominant structure present at the isoelectric point? COOH H C H2 N NH2 Figure 1.10. The structure of lysine. (Answers to problems can be found on pp. 277–279.) 2 Partition coefficient and biopharmacy When a substance (or solute) is added to a pair of immiscible solvents, it distributes itself between the two solvents according to its affinity for each phase. A polar compound (e.g. a sugar, amino acid or ionised drug) will tend to favour the aqueous or polar phase, whereas a non-polar compound (e.g. an unionised drug) will favour the non-aqueous or organic phase. The added substance distributes itself between the two immiscible solvents according to the partition law, which states that ‘a given substance, at a given temperature, will partition itself between two immiscible solvents in a constant ratio of concentrations’. This constant ratio is called the partition coefficient of the substance, and may be expressed mathematically as [organic] P= (2.1) [aqueous] where P is the partition coefficient of the substance; [organic] is the con- centration of substance in the organic, or oil phase; and [aqueous] is the concentration of substance in the water phase. As an example, consider the distribution of 100 mg of a drug between 50 mL of an organic solvent (e.g. ether, chloroform or octanol) and 50 mL of water. The drug is added to the two immiscible solvents in a separating funnel and allowed to equilibrate. When the organic layer is analysed, it is found to contain 66.7 mg of compound. From these data the partition coefficient and the percentage of the drug extracted into the organic layer can be calculated (see Fig. 2.1). The mass of drug in the water phase = 100 − 66.7 mg = 33.3 mg; the concentration of drug in the organic phase = 66.7/50 = 1.33 mg mL −1 , and the concentration of drug in the water phase = 33.3/50 = 0.67 mg mL −1. Therefore, the partition coefficient is given by [organic] 1.33 mg mL −1 = =2 [aqueous] 0.67 mg mL −1 30 Essentials of Pharmaceutical Chemistry Organic (50 mL) Water (50 mL) [D]org 66.7 mg (2) 33.3 mg (1) [D]aq (unionised) (unionised) Figure 2.1. Simple partition law. The partition coefficient is a ratio of concentrations, so the units cancel and P has no units. The percentage of drug extracted in the above example is simply given by the mass of drug in the organic phase divided by the total mass of drug, i.e. 66.7/100 = 66.7%. The partition coefficient is an important piece of information as it can be used to predict the absorption, distribution and elimination of drugs within the body. Knowledge of the value of P can be used to predict the onset of action of drugs or the duration of action of drugs, or to tell whether a drug will be active at all. Part of medicinal chemistry, the science of rational drug design, involves structure–activity relationships, where the partition coefficient is used in mathematical equations that try to relate the biological activity of a drug to its physical and chemical characteristics. In case this sounds too much like an advert for the partition coefficient, in reality the simple relationship above only applies if the solute in question does not ionise at the pH of measurement. If the solute is a weak acid or weak base (and a huge number of drugs are), then ionisation to form an anion or a cation will considerably alter the solubility profile of the drug. A fully ionised species will be much more soluble in water than the unionised acid or base, and so the above ratio will vary depending on the pH at which the measurement was carried out. There are two ways round this problem: either the experimental condi- tions are adjusted to ensure that the measured P is the partition coefficient of the unionised molecule (this means that the P value for acids is measured at low pH when the acid is unionised and, similarly, the partition coefficient of a base is measured at high pH to prevent ionisation); or, better, the ratio above is redefined as the apparent partition coefficient, to differentiate it  Partition coefficient and biopharmacy 31 from the partition coefficient of the unionised species, which is now termed the true partition coefficient. The apparent partition coefficient ( P app ) is dependent on the proportion of substance present in solution, which in turn depends upon the pH of the solution, or P app = P × f unionised (2.2) where f unionised equals the fraction of the total amount of drug unionised at that pH. It follows that if f unionised = 1 then P app = P true and the compound is unionised. To illustrate the effect of ionisation, consider again the drug in the example above. If the pH of the aqueous phase is adjusted so that the drug becomes 66.7% ionised, only 40 mg of the drug partitions into the organic phase (since the ionised drug will be less soluble in the organic solvent), and the partition coefficient can be recalculated (see Fig. 2.2). Organic (50 mL) Aqueous (50 mL) 40 mg (2) 20 mg (1) (unionised) (unionised) 40 mg (2) (ionised) Figure 2.2. The partition of ionised drug. The mass of drug in the water phase = 100 − 40 = 60 mg. The mass of ionised drug in the water phase = total mass×fraction ionised, which is 60 × 0.666 = 40 mg. The mass of unionised drug in the water phase = 60 × 0.333 = 20 mg. The concentration of drug in the organic phase = 40/50 = 0.8 mg mL −1. The concentration of unionised drug in the water phase = 20/50 = 0.4 mg mL −1. The concentration of total drug in the water phase = 60/50 = 1.2 mg mL −1. The percentage of drug extracted into the organic phase = (40 mg/100 mg) × 100 = 40%. The partition coefficient of the unionised drug (the true partition coeffi- cient) should remain constant and is given by [drug] in organic phase P= [unionised drug] in water 32 Essentials of Pharmaceutical Chemistry 0.8 mg L −1 P= =2 0.4 mg mL −1 the same answer as obtained above. Using the total concentration of drug in the aqueous phase allows the apparent partition coefficient to be calculated: [drug] in organic phase P app = total [drug] in aqueous phase 0.8 mg mL −1 P app = 1.2 mg mL −1 P app = 0.67 The answer for Papp can be checked by use of Eq. (2.2): P app = P × f unionised 0.67 = 2 × 0.33 The range of possible values of P found in drug molecules is huge, from small fractions through to values of several thousand. For this reason, it is common to quote the logarithm (to the base 10) of the partition coefficient, or log P. This is particularly true in quantitative structure–activity relation- ships (QSAR), where the physicochemical properties of a drug (such as hydrophobicity, steric interactions or electronic effects) are quantified and an equation is derived that can be used to predict the biological activity of other, similar drugs. The technique of QSAR became popular with the ad- vent of powerful computers able to handle the multiple regression analysis necessary to obtain the quite complex equations required. A detailed study of QSAR is beyond the scope of this book, but more advanced textbooks of medicinal chemistry contain many examples of the ability of QSAR equations to predict biological activity. Experimental measurement of the partition coefficient There are three convenient ways in which P can be determined in the chemistry laboratory. These are the original shake flask method, the use of thin-layer chromatography or the use of reversed-phase, high-performance liquid chromatography. Shake flask method In the shake flask method, the drug whose P is to be determined is traditionally added to a separating funnel containing the two immiscible phases, although it works just as well to use a centrifuge tube (and requires  Partition coefficient and biopharmacy 33 less sample). The two immiscible phases chosen are usually 1-octanol and pH 7.4 buffer. Octanol is used in partition coefficient work because the answers obtained from octanol seem to correlate best with biological data obtained in vivo. This may be because the eight carbon atoms are essentially hydrophobic (or water-hating) and the one hydroxyl group is hydrophilic (water-loving) and together they give the closest balance to that found in human cell membranes. The aqueous buffer at pH 7.4 represents aqueous compartments within the body, e.g. blood plasma. The two phases are thoroughly mixed to give buffer-saturated octanol in the top phase and octanol-saturated buffer in the bottom. Once the two phases have separated (this can take a while), the drug is added and the whole flask is shaken mechanically for at least an hour. The two phases are allowed to separate (or centrifuged, if you are in a hurry) and the concentration of drug in the aqueous phase is then determined. This may be done by titration if the drug is sufficiently acidic or basic or, more usually, spectrophotometrically. The concentration in the octanol phase is found by subtraction and the value of P is calculated. This method works perfectly well if there is sufficient sample and the drug possesses a chromophore to allow spectroscopic assay of the aqueous phase. What is important in liquid–liquid extractions of this type is not the volume of the organic phase but rather the number of times the extraction is carried out. Five extractions of 10 mL organic phase will remove more compound than one extraction of 50 mL, even though the total volume of organic solvent used is the same. Similarly, ten extractions of 5 mL will be more efficient still, and so on. This effect (which is general to all extractions) is obvious when thought about. Each time one phase is removed and replaced by fresh solvent, the equilibrium for the partitioning process must re-establish according to the partition coefficient ratio and drug must leave the aqueous phase to enter the organic phase and restore the equilibrium ratio. An equation can be derived to calculate the increase in efficiency of multiple extractions versus one single extraction:  n A Wn = W (2.3) PS + A where W n is the mass of drug remaining in the aqueous phase after n extractions, W is the initial mass of drug in the aqueous phase, A is the volume of the aqueous phase, S is the volume of solvent (or organic) phase, P is the partition coefficient and n is the number of extractions. Equation (2.3) is derived as follows: [organic] P= [aqueous] 34 Essentials of Pharmaceutical Chemistry or, using the terms defined above, (W − W 1 ) / S P= (W 1 /A ) Therefore, (W − W 1 ) A P= × W1 S or PS (W − W 1 ) W = = −1 A W1 W1 Hence, W PS PS A (PS + A) = +1= + = W1 A A A A Therefore, the fraction of drug remaining in the aqueous phase is W1 A = (2.4) W (PS + A) This expression is valid for one extraction; it follows that if the extraction is repeated n times, the overall expression is simply given by Eq. (2.4) repeated n times, which, with subscript 1 replaced by n, is Eq. (2.3). Thin-layer chromatography (TLC) In this technique, the R f value of the drug is related mathematically to the partition coefficient. A thin-layer plate, or a paper sheet, is pre-coated with organic phase (usually paraffin or octanol) and allowed to dry. Sample is applied to the origin and the plate is allowed to develop. The mobile phase used is either water or a mixture of water and a miscible organic solvent (such as acetone) to improve the solubility of the drug. Once the plate has developed, the spots are visualised (using an ultra- violet lamp if the drug possesses a chromophore, or iodine vapour if it does not) and the R f for each spot is determined. The R f is the distance moved by the spot divided by the distance moved by the solvent front, and is expressed as a decimal. The R f can be related to the partition coefficient by equations of the type k P= (2.5) (1/R f ) − 1 where k is a constant for the given system, which is determined by running a number of standard compounds of known P in the system and calculating k.  Partition coefficient and biopharmacy 35 The TLC method of determining P works best for compounds of similar structure and physical properties. The advantages of using this technique to determine P are that many compounds can be run simultaneously on one plate, and very little sample is required. On the other hand, finding suitable standards can be difficult, and mobile phases containing a large amount of aqueous solvent may take many hours to run up a large TLC plate. High-performance liquid chromatography (HPLC) This method of analysis relies on the same chemical principles as the deter- mination by TLC, except that the efficiency (and the cost) of the technique has increased greatly. Instead of the R f value, the retention time of the drug is measured and related to P by equations similar to Eq. (2.5) for TLC. The retention time, as its name suggests, is the time taken for the sample to elute from the HPLC column. The major drawback with using this technique to determine P is detecting the drug if it does not possess a chromophore, when a UV detector cannot be used. In cases like this, use must be made of an HPLC system connected to a refractive index (RI) detector or an electrochemical detector (ECD). A RI detector relies on changes in the refractive index of the mobile phase as a solute elutes to detect a signal, while an ECD functions like a little electrode to oxidise or reduce the analyte as it elutes. In either case, before the determination of P is carried out, you should seriously consider measuring P for another drug! My PhD supervisor had a saying: ‘Never make a compound you cannot name’; to that can be added the advice ‘Never make a compound that cannot be detected by a UV detector’. Many entertaining hours can be spent optimising HPLC systems with RI or ECD, but if you want to finish before your children grow up, these methods of detection are best avoided. There are some advantages to the HPLC method of determining P, namely that HPLC does not require much sample and that the sample does not have to be 100% pure. Also, once the complete system has been obtained, the cost of the determination is limited to the purchase of HPLC-grade solvents and electricity. Drug absorption, distribution and bioavailability The study of the fate of a drug administered to an organism is called pharma- cokinetics. This discipline involves measuring or predicting the absorption, distribution, metabolism and excretion (usually known by the acronym ADME) of the drug in the body. Pharmacokinetics has been described as ‘what the body does to the drug’ as opposed to pharmacodynamics, which is the study of mechanisms of drug action and the biochemical changes brought about by treatment with the drug or ‘what the drug does to the 36 Essentials of Pharmaceutical Chemistry body’. Some older textbooks use the expression ‘molecular pharmacology’ instead of pharmacodynamics. Bioavailability (symbol F) is a measure of the extent to which a drug reaches the bloodstream and is available at its site of action. The bioavail- ability of a drug administered by intra-venous (i.v.) injection is defined as 1 (since the entire dose is available in the systemic circulation). Problems start to appear, however, if the drug is administered by a non-parenteral route (e.g. oral, rectal, topical). In these cases, the bioavailability of a drug is often considerably less than 1 due to a number of factors, such as poor absorption from the gut (in the case of an oral medicine), extensive binding to circulating plasma proteins, or rapid ‘first pass’ metabolism in the liver. The most popular method of administering drugs and medicines, at least in the UK, is the oral route. Tablets, capsules or oral liquids are swallowed and, once in the stomach, the tablet or capsule disintegrates to release the active drug molecule. Interestingly, a drug is not considered to be in the body until it has been absorbed across the gut wall and into the bloodstream. The gut can be thought of as a hollow tube running through the body, open at both ends (hopefully not at the same time) and, as such, the gut contents are considered outside the body. Passage into the body must be achieved by absorption across a biological membrane; for the oral route of drug administration, this is the cell membrane of cells lining the wall of the stomach and the intestine. Once the drug has passed through the gut membrane into the bloodstream, it then has to travel to its site of action and diffuse out of the bloodstream to the receptor on some, perhaps distant, cell membrane. In the case of drugs acting on the brain or spinal cord (the central nervous system, or CNS) the drug must partition across the blood–brain barrier to gain access to the CNS. The blood–brain barrier is, in reality, the cell membranes of glial cells (or astrocytes) lining the blood vessels within the brain. These cells fuse together very closely to form a tight, high-resistance ‘lipid barrier’ that restricts the passage of many drug molecules, especially if the drug molecules are polar. It has been estimated that the blood–brain barrier prevents the brain uptake of >98% of all potential neurotherapeutics. This barrier is designed to protect the delicate structures of the brain from damage by harmful compounds that may gain access to the bloodstream, but it can be a problem for drug administration. Some infectious diseases, such as malaria, can spread to the brain and, once established, can be very difficult to treat, since drugs used to eradicate the infection in other parts of the body cannot cross the blood–brain barrier to get at the infection in the CNS. This creates a ‘reservoir of infection’ which can re-infect the rest of the body after treatment. A similarly depressing picture exists with tumours in the brain. Con- ventional anticancer chemotherapy often cannot penetrate the barrier to attack the tumour. The CNS, however, does require low-molecular-weight  Partition coefficient and biopharmacy 37 molecules to grow and function and these small polar molecules (e.g. amino acids, sugars) have their own transport proteins located at the blood–brain barrier that act to transfer the essential compound through the barrier in a process called carrier-mediated transport. Biological membranes vary in structure and function throughout the body, but there are some common structural features and properties (see Fig. 2.3). A cell membrane is composed of a bilayer of fatty molecules known as phospholipids. These compounds are amphoteric in nature, pos- sessing a non-polar region of hydrocarbon chains that are buried inside the cell membrane, and a polar region comprising negatively charged phos- phoric acid head groups. These ionised groups are exposed to the aqueous surroundings of the extracellular and intracellular fluids of the cell. The cell membrane has to be fatty and non-polar in nature to allow it to successfully separate the aqueous compartments of the body. Buried within this lipid bilayer are large globular protein molecules. These macromolecules function as ion channels (e.g. the Na + channel of nerve membranes), transmembrane receptors (like the β adrenoceptor) or transport proteins (as in the electron transport chain of mitochondria). Human cell membranes also contain high concentrations of the steroid cholesterol, particularly in nerve tissue. Chemically, cholesterol is a cyclopentanoperhydrophenanthrene derivative, but it is much simpler to use the trivial name and call this important group of compounds ‘steroids’. The structure of cholesterol and a general structure of membrane phospholipids are shown in Fig. 2.4. Protein Polar head Hydrocarbon chains Phospholipid bilayer Figure 2.3. A fluid mosaic model of a cell membrane. Cholesterol gets a bad press nowadays. Tabloid newspapers and tele- vision programmes seem to have latched onto cholesterol as the villain of a healthy lifestyle. It is true that high levels of cholesterol in the diet, coupled with high salt intake and lack of exercise, are blamed for causing coronary heart disease and strokes. In cell membranes, however, cholesterol increases membrane rigidity and is essential for maintaining the integrity of the membrane — without cholesterol your cells would leak. Other important steroids include the male sex hormone, testosterone, and the female sex 38 Essentials of Pharmaceutical Chemistry O H3C CH3 CH3 CH2O C R1 H O CH3 CH3 H C O C R2 O H H HO CH2 O P R3 H OH Figure 2.4. The structures of cholesterol and phospholipids. R1 and R2 = palmityl, stearyl or oleyl. R3 = ethanolamine, choline, serine, inositol or glycerol. hormones oestrogen and progestogen in addition to drugs such as digoxin and beclometasone. The lipid bilayer of the cell membrane presents a significant barrier to drug transport and for a small drug molecule to travel across membranes, one of two things must happen: either the drug must cross the membrane by passive diffusion down its concentration gradient or the drug has to be transported across the membrane, against the concentration gradient, with the expenditure of energy, a process called active transport. Passive diffusion Passive diffusion is probably the most important mechanism by which small drug molecules gain access to the body. The drug molecule must be in solution and it partitions into the lipophilic cell membrane, diffuses across the cell and then partitions out of the cell and into the aqueous compartment on the other side. Drugs that are very lipid soluble (such as the antifungal agent griseofulvin) are so water insoluble that they partition into the cell membrane but then stick in the lipid membrane and do not partition out of the membrane and into the aqueous compartments inside the cell. Similarly, drugs that are very water soluble will not partition well into a non-polar lipid membrane and will tend to stay in the aqueous contents of the gut, or if they do manage to cross the gut membrane will stick in the aque- ous intracellular solution. Clearly, for a drug to be successfully absorbed from the gut it must possess an intermediate level of water solubility and lipid solubility: a sort of ‘Goldilocks effect’ whereby the drug is not too hydrophobic, not too hydrophilic, but possesses just the right degree of solubility to partition through biological membranes. In general, drugs that are strongly acidic with a pK a < 2 or strongly basic (pK a > 10) will not cross membranes very well since they will be >99.99% ionised at the pH values found in the gut.  Partition coefficient and biopharmacy 39 Transfer across membranes occurs down a concentration gradient (i.e. from regions of high drug concentration to regions where the concentration is lower). The process of diffusion can be described by Fick’s law(named after the German physiologist, Adolf Fick), which states dm P DA( C 2 − C 1 ) = dt d where dm /dt is the rate of appearance of drug within the cell (or rate of transfer), P is the partition coefficient of the drug, D is a diffusion coeffi- cient for the membrane, A is the surface area of membrane available for absorption, C 2 and C 1 are the concentrations of drug on the external and internal surfaces, respectively, and d is the thickness of the cell membrane. Rate of diffusion is favoured by high values of P, high membrane surface area and a steep concentration gradient across a thin membrane. Passive diffusion can only occur with small molecules (e.g. drugs with relative molecular masses of approximately 1000 or less). This excludes large